# Tetration Forum

Full Version: An alternate power series representation for ln(x)
You're currently viewing a stripped down version of our content. View the full version with proper formatting.
This proof involves the use of a new operator:
$x \bigtriangleup y = ln(e^x + e^y)$

and it's inverse:
$x \bigtriangledown y = ln(e^x - e^y)$

and the little differential operator:
$\bigtriangleup \frac{d}{dx} f(x) = \lim_{h\to\ -\infty} [f(x \bigtriangleup h) \bigtriangledown f(x)] - h$

see here for more: http://math.eretrandre.org/tetrationforu...40#pid5740

I'll use these specifically:
$\bigtriangleup \frac{d}{dx} [f(x) \bigtriangleup g(x)] =(\bigtriangleup \frac{d}{dx} f(x)) \bigtriangleup (\bigtriangleup \frac{d}{dx} g(x))$
and
$\bigtriangleup \frac{d}{dx} xn = x(n-1) +ln(n)$

The proof starts out by first proving:
$\bigtriangleup \frac{d}{dx} e^x = e^x$

first give the power series representation of e^x
$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$

And given: $ln(x + y) = ln(x) \bigtriangleup ln(y)$

We take the ln of e^x to get an infinite series of deltations, if:
$\bigtriangleup \sum_{n=0}^{R} f(n) = f(0) \bigtriangleup f(1) \bigtriangleup ... f®$ represents a series of deltations

then
$x = \bigtriangleup \sum_{n=0}^{\infty} nln(x) - ln(n!)$

and therefore if we let x = e^x

$e^x = \bigtriangleup \sum_{n=0}^{\infty} nx - ln(n!)$

and now we have an infinite lowered polynomial which when little differentiated equals itself. It's the little polynomial equivalent of e^x's perfect taylor series.

therefore:
$\bigtriangleup \frac{d}{dx} e^x = e^x$

and using the chain rule:
$\bigtriangleup \frac{d}{dx} f(g(x)) = f'(g(x)) + g'(x)$ where f'(x) is taken to mean the little derivative of f(x).

we get the result:
$\bigtriangleup \frac{d}{dx} ln(x) = -x$

and now we can solve for the k'th little derivative of ln(x) using the little power rule, k E N

$\bigtriangleup \frac{d^k}{dx^k} ln(x) = -kx + ln((-1)^{k-1} (k-1)!)$

and so if the little derivative Taylor series is given by:
$f(x) = \bigtriangleup \sum_{n=0}^{\infty} f^{(n)}(a) + n(x \bigtriangledown a) - ln(n!)$
where $f^{(n)}(x)$ is the n'th little derivative of $f$.

we can take the little derivative taylor series of ln(x) centered about 1.
The first term is equal to 0, so I'll start the series from n = 1.

Here it is in two steps:
$ln(x) = 0 \bigtriangleup (\bigtriangleup \sum_{n=1}^{\infty} n(x \bigtriangledown 1) + ln((-1)^{n-1}(n-1)!)-ln(n!)-n)$

$ln(x) = 0 \bigtriangleup (\bigtriangleup \sum_{n=1}^{\infty} ln(\frac{(-1)^{n-1}}{n}) + n(x \bigtriangledown 1) - n)$

now let's plug this in our formula for $x \bigtriangleup y$; it works for an infinite sum because $\bigtriangleup$ is commutative and associative.

$ln(x) = ln(1 + \sum_{n=1}^{\infty} e^{ln(\frac{(-1)^{n-1}}{n}) + n(x \bigtriangledown 1)-n})$

now since:
$x \bigtriangledown 1 = ln(e^x - e)$

$ln(x) = ln(1 + \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{ne^n}(e^x - e)^n)$

now take the lns away and

$x = 1 + \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{ne^n}(e^x - e)^n$

and now if we let x = ln(x) we get:
$ln(x) = 1 + \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{ne^n}(x - e)^n$

And there, that's it. I haven't been able to check if this series converges, I don't have many convergence tests. I'm just sure that the algebra is right. I haven't tried computing it. In my gut it doesn't look like it will converge, but I have faith.
Doing some tests it converges for 1, 2, 4, 5 after about 10 terms, it fails to converge for 6. Still, I'm enthralled this works.

Doing some more tests, if I do the same process over again, but take the little taylor series of ln(x) centered about 0 instead of 1 (we can do this because negative infinity is the identity of $\bigtriangleup$) I find the normal taylor series of ln(x): $ln(x) = \sum_{n=1}^{\infty} (x-1)^n \frac{(-1)^{n+1}}{n}$

so this leads me to believe that if we take the little taylor series centered about K and then do the same process to get a normal power series expression, and we let K tend to infinity, we will have an ever growing radius of convergence. That is since doing this method centered about 1 has a bigger radius of convergence than doing this method centered about 0.
(05/07/2011, 08:41 PM)JmsNxn Wrote: [ -> ]This proof involves the use of a new operator:
$x \bigtriangleup y = ln(e^x + e^y)$

and it's inverse:
$x \bigtriangledown y = ln(e^x - e^y)$

and the little differential operator:
$\bigtriangleup \frac{d}{dx} f(x) = \lim_{h\to\ -\infty} [f(x \bigtriangleup h) \bigtriangledown f(x)] - h$

But your operator can be expressed with the classical differentiation, see:

$\begin{eqnarray}
\bigtriangleup \frac{d}{dx} f(x) &=& \lim_{h\to\ -\infty} [f(x \bigtriangleup h) \bigtriangledown f(x)] - h\\
&=& \lim_{h\to\ -\infty} \quad\ln[\exp(f(x \bigtriangleup h)) - \exp(f(x))] - h\\
&=& \ln\quad\lim_{d\to 0} \frac{\exp(f(x \bigtriangleup \log(d))) - \exp(f(x))}{d}\\
&=& \ln\quad\lim_{d\to 0} \frac{\exp(f(\ln(e^x + d)) - \exp(f(x))}{d}\\
&=& \ln\quad\lim_{d\to 0} \frac{\exp(f(\ln(e^x + d)) - \exp(f(\ln(\exp(x)))))}{d}\\
& =& \ln((\exp\circ f\circ \ln)'(\exp(x)))
\end{eqnarray}$

Or purely functional with the composition operation $\circ$: $\bigtriangleup \frac{d}{dx} f = \ln\circ(\exp\circ f\circ \ln)'\circ\exp$

PS: when you write ln with backslash in front:
Code:
$$\ln(x)$$
you get a better ln-typesetting.
Yeah, I was aware of a direct relation to differentiation:

$\frac{d}{dx} f(x) = e^{(\bigtriangleup \frac{d}{dx} f(x)) + x - f(x)}$

And I've extended the definition of the series to:

$\ln(x) = a + \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{ne^{an}}(x-e^a)^n$

And I've found a beautiful return:

$\ln(x) = e + \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{ne^{ne}}(x-e^e)^n$

Converges for values as high as 30. My computer overflows before it stops converging.

I think my assumptions were correct. If this is true, I might find an infinite convergent power series for ln(x)

edit:
sadly, doesn't converge for values less than 1
(05/07/2011, 11:20 PM)JmsNxn Wrote: [ -> ]$\ln(x) = a + \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{ne^{an}}(x-e^a)^n$

But this follows directly from the definition of the logarithm, by the following equivalent transformations:
$\begin{eqnarray}
\ln(x) &=& a + \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{ne^{an}}(x-e^a)^n\\
&=& a + \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}(x/e^a-1)^n\\
\ln(e^a y) &=& a + \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}(y-1)^n\\
a + \ln(y) &=& a + \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}(y-1)^n\\
\ln(y) &=& \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}(y-1)^n
\end{eqnarray}$

Quote:sadly, doesn't converge for values less than 1
The $\ln$ series converges for $|y-1|<1$, hence your series converges for $|x/e^a-1|<1$, this should include all values $x\in (0,2e^a)$?
I'm sorry to ask, but how did you get from step 1 to step 2? I don't understand where the $e^{an}$ went to in the denominator. Otherwise, though, that's a nice proof.
(05/08/2011, 07:54 PM)JmsNxn Wrote: [ -> ]I'm sorry to ask, but how did you get from step 1 to step 2? I don't understand where the $e^{an}$ went to in the denominator. Otherwise, though, that's a nice proof.

Oh thats just:
$\frac{1}{e^{an}}(x-e^a)^n=\frac{(x-e^a)^n}{(e^a)^n} = \left(\frac{x-e^a}{e^a}\right)^n = \left(\frac{x}{e^a} - 1\right)^n$
(05/08/2011, 08:28 PM)bo198214 Wrote: [ -> ]Oh thats just:
$\frac{1}{e^{an}}(x-e^a)^n=\frac{(x-e^a)^n}{(e^a)^n} = \left(\frac{x-e^a}{e^a}\right)^n = \left(\frac{x}{e^a} - 1\right)^n$

Oh that's so simple! Thanks for the help.